3.528 \(\int \frac{(a+b x^2)^{3/2} (A+B x^2)}{x^2} \, dx\)

Optimal. Leaf size=109 \[ \frac{x \left (a+b x^2\right )^{3/2} (a B+4 A b)}{4 a}+\frac{3}{8} x \sqrt{a+b x^2} (a B+4 A b)+\frac{3 a (a B+4 A b) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{8 \sqrt{b}}-\frac{A \left (a+b x^2\right )^{5/2}}{a x} \]

[Out]

(3*(4*A*b + a*B)*x*Sqrt[a + b*x^2])/8 + ((4*A*b + a*B)*x*(a + b*x^2)^(3/2))/(4*a) - (A*(a + b*x^2)^(5/2))/(a*x
) + (3*a*(4*A*b + a*B)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(8*Sqrt[b])

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Rubi [A]  time = 0.0411167, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {453, 195, 217, 206} \[ \frac{x \left (a+b x^2\right )^{3/2} (a B+4 A b)}{4 a}+\frac{3}{8} x \sqrt{a+b x^2} (a B+4 A b)+\frac{3 a (a B+4 A b) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{8 \sqrt{b}}-\frac{A \left (a+b x^2\right )^{5/2}}{a x} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^(3/2)*(A + B*x^2))/x^2,x]

[Out]

(3*(4*A*b + a*B)*x*Sqrt[a + b*x^2])/8 + ((4*A*b + a*B)*x*(a + b*x^2)^(3/2))/(4*a) - (A*(a + b*x^2)^(5/2))/(a*x
) + (3*a*(4*A*b + a*B)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(8*Sqrt[b])

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^2} \, dx &=-\frac{A \left (a+b x^2\right )^{5/2}}{a x}-\frac{(-4 A b-a B) \int \left (a+b x^2\right )^{3/2} \, dx}{a}\\ &=\frac{(4 A b+a B) x \left (a+b x^2\right )^{3/2}}{4 a}-\frac{A \left (a+b x^2\right )^{5/2}}{a x}+\frac{1}{4} (3 (4 A b+a B)) \int \sqrt{a+b x^2} \, dx\\ &=\frac{3}{8} (4 A b+a B) x \sqrt{a+b x^2}+\frac{(4 A b+a B) x \left (a+b x^2\right )^{3/2}}{4 a}-\frac{A \left (a+b x^2\right )^{5/2}}{a x}+\frac{1}{8} (3 a (4 A b+a B)) \int \frac{1}{\sqrt{a+b x^2}} \, dx\\ &=\frac{3}{8} (4 A b+a B) x \sqrt{a+b x^2}+\frac{(4 A b+a B) x \left (a+b x^2\right )^{3/2}}{4 a}-\frac{A \left (a+b x^2\right )^{5/2}}{a x}+\frac{1}{8} (3 a (4 A b+a B)) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )\\ &=\frac{3}{8} (4 A b+a B) x \sqrt{a+b x^2}+\frac{(4 A b+a B) x \left (a+b x^2\right )^{3/2}}{4 a}-\frac{A \left (a+b x^2\right )^{5/2}}{a x}+\frac{3 a (4 A b+a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{8 \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.190682, size = 87, normalized size = 0.8 \[ \frac{1}{8} \sqrt{a+b x^2} \left (\frac{3 \sqrt{a} (a B+4 A b) \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{b} \sqrt{\frac{b x^2}{a}+1}}-\frac{8 a A}{x}+5 a B x+4 A b x+2 b B x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^(3/2)*(A + B*x^2))/x^2,x]

[Out]

(Sqrt[a + b*x^2]*((-8*a*A)/x + 4*A*b*x + 5*a*B*x + 2*b*B*x^3 + (3*Sqrt[a]*(4*A*b + a*B)*ArcSinh[(Sqrt[b]*x)/Sq
rt[a]])/(Sqrt[b]*Sqrt[1 + (b*x^2)/a])))/8

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Maple [A]  time = 0.008, size = 125, normalized size = 1.2 \begin{align*}{\frac{Bx}{4} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,Bax}{8}\sqrt{b{x}^{2}+a}}+{\frac{3\,{a}^{2}B}{8}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){\frac{1}{\sqrt{b}}}}-{\frac{A}{ax} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{Abx}{a} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,Abx}{2}\sqrt{b{x}^{2}+a}}+{\frac{3\,Aa}{2}\sqrt{b}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(3/2)*(B*x^2+A)/x^2,x)

[Out]

1/4*x*B*(b*x^2+a)^(3/2)+3/8*B*a*x*(b*x^2+a)^(1/2)+3/8*B*a^2/b^(1/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))-A*(b*x^2+a)^
(5/2)/a/x+A*b/a*x*(b*x^2+a)^(3/2)+3/2*A*b*x*(b*x^2+a)^(1/2)+3/2*A*b^(1/2)*a*ln(x*b^(1/2)+(b*x^2+a)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.6328, size = 424, normalized size = 3.89 \begin{align*} \left [\frac{3 \,{\left (B a^{2} + 4 \, A a b\right )} \sqrt{b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) + 2 \,{\left (2 \, B b^{2} x^{4} - 8 \, A a b +{\left (5 \, B a b + 4 \, A b^{2}\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{16 \, b x}, -\frac{3 \,{\left (B a^{2} + 4 \, A a b\right )} \sqrt{-b} x \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) -{\left (2 \, B b^{2} x^{4} - 8 \, A a b +{\left (5 \, B a b + 4 \, A b^{2}\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{8 \, b x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^2,x, algorithm="fricas")

[Out]

[1/16*(3*(B*a^2 + 4*A*a*b)*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(2*B*b^2*x^4 - 8*A*a*
b + (5*B*a*b + 4*A*b^2)*x^2)*sqrt(b*x^2 + a))/(b*x), -1/8*(3*(B*a^2 + 4*A*a*b)*sqrt(-b)*x*arctan(sqrt(-b)*x/sq
rt(b*x^2 + a)) - (2*B*b^2*x^4 - 8*A*a*b + (5*B*a*b + 4*A*b^2)*x^2)*sqrt(b*x^2 + a))/(b*x)]

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Sympy [B]  time = 8.71008, size = 216, normalized size = 1.98 \begin{align*} - \frac{A a^{\frac{3}{2}}}{x \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{A \sqrt{a} b x \sqrt{1 + \frac{b x^{2}}{a}}}{2} - \frac{A \sqrt{a} b x}{\sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 A a \sqrt{b} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{2} + \frac{B a^{\frac{3}{2}} x \sqrt{1 + \frac{b x^{2}}{a}}}{2} + \frac{B a^{\frac{3}{2}} x}{8 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 B \sqrt{a} b x^{3}}{8 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 B a^{2} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{8 \sqrt{b}} + \frac{B b^{2} x^{5}}{4 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)*(B*x**2+A)/x**2,x)

[Out]

-A*a**(3/2)/(x*sqrt(1 + b*x**2/a)) + A*sqrt(a)*b*x*sqrt(1 + b*x**2/a)/2 - A*sqrt(a)*b*x/sqrt(1 + b*x**2/a) + 3
*A*a*sqrt(b)*asinh(sqrt(b)*x/sqrt(a))/2 + B*a**(3/2)*x*sqrt(1 + b*x**2/a)/2 + B*a**(3/2)*x/(8*sqrt(1 + b*x**2/
a)) + 3*B*sqrt(a)*b*x**3/(8*sqrt(1 + b*x**2/a)) + 3*B*a**2*asinh(sqrt(b)*x/sqrt(a))/(8*sqrt(b)) + B*b**2*x**5/
(4*sqrt(a)*sqrt(1 + b*x**2/a))

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Giac [A]  time = 1.17245, size = 154, normalized size = 1.41 \begin{align*} \frac{2 \, A a^{2} \sqrt{b}}{{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a} + \frac{1}{8} \,{\left (2 \, B b x^{2} + \frac{5 \, B a b^{2} + 4 \, A b^{3}}{b^{2}}\right )} \sqrt{b x^{2} + a} x - \frac{3 \,{\left (B a^{2} \sqrt{b} + 4 \, A a b^{\frac{3}{2}}\right )} \log \left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2}\right )}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^2,x, algorithm="giac")

[Out]

2*A*a^2*sqrt(b)/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a) + 1/8*(2*B*b*x^2 + (5*B*a*b^2 + 4*A*b^3)/b^2)*sqrt(b*x^2
+ a)*x - 3/16*(B*a^2*sqrt(b) + 4*A*a*b^(3/2))*log((sqrt(b)*x - sqrt(b*x^2 + a))^2)/b